STAT 305 Homework # 4 Solution

Show all of your work on this assignment and answer each question fully in the given context.

Please staple your assignment!

1. Chapter 4, Section 1, Exercise 3 (unless directed otherwise you may use JMP; include plotsas requested) (page 140) [5 pts each, 25 pts total]

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STAT 305 Homework # 4 Solution

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STAT 305 Homework # 4 Solution

2. Chapter 4, Section 1, Exercise 4 (unless directed otherwise you may use JMP; include plotsas requested) (page 140) [5 pts each, 15 pts total]

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STAT 305 Homework # 4 Solution

3. Chapter 4, Section 2, Exercise 1 (unless directed otherwise you may use JMP; include plotsas requested) (page 161) [10 pts]

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STAT 305 Homework # 4 Solution

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STAT 305 Homework # 4 Solution

4. Chapter 4, Section 2, Exercise 2 (unless directed otherwise you may use JMP; include plotsas requested; skip part h) (page 161) [5 pts each, 35 pts total]

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STAT 305 Homework # 4 Solution

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STAT 305 Homework # 4 Solution

5. The major cause of axel failure in freight trucks is when shippers exceed the recommendedweight limits that can be handled by the axels. Issues resulting from these failures havebeen becoming more frequent as shippers try to cut corners, leading members of the state’sDepartment of Transportation to ask one of their civil engineers to look into the availabledata and better advise them on the relationship between excessive weight and axel failure.

A company manufacturing axels provides the engineer with data gathered from conductingexperiments loading axels with excessive weight and simulating traveling conditions. Thedata consists of two columns, excessive weight (in tonnes) is the amount of weight overthe limit that was placed on the axel, and distance to failure (in tens of thousands ofmiles) is the simulated distance to the axel’s failure.

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STAT 305 Homework # 4 Solution

0

20

40

60

0.0 0.5 1.0 1.5 2.0 2.5Weight Exceeding Guidelines (in tonnes)

Trav

el D

ista

nce

to F

ailu

re (

10,0

00 m

iles)

Here are some summaries of the data:

6. Using the summaries above, fit a linear relationship between weight exceeding guidelines(x) and travel distance to failure (y).[10 pts]

The fitted line equation isy = b0 + b1 · x

We can use the information above to get the value for b1 and b0:

b1 =

∑ni=1 xiyi − nxy∑ni=1 x

2i − nx2

=(1982) − (50)(64/50)(1996/50)

107 − 50(64/50)2

= −22.8421052631579

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STAT 305 Homework # 4 Solution

and with b1 we can find the value fo b0:

b0 = y − b1x

= (1996/50) − (−22.8421052631579)(64/50)

= 69.1578947368421

Which gives us the fitted equation of

y = 69.15 − 22.84 · x

7. Write the equation of the fitted linear relationship. [5 pts]

y = 69.15 − 22.84 · x

8. Find and interpret the value of R2 for the fitted linear relationship.[5 pts]

Since we are using a linear relationship, we can get R2 from r:

r =

∑ni=1 xiyi − nxy√(∑n

i=1 x2i − nx2

) (∑ni=1 y

2i − ny2

)=

(1982) − (50)(64/50)(1996/50)√(107 − 50(64/50)2) (94078 − (50)(1996/50)2)

= −0.953352777638285

So R2 = (r)2 = 0.908881518630634

This means that 93.00% of our the variablity in travel distance to failure can be explained bythe linear relationship with weight exceeding guidelines.

9. Using the fitted line, provide a predicted value of travel distance to failure when the weightexceeding the guidelines is 3.4 tonnes.[5 pts]

y = 69.15 − 22.84(3.4) = −8.50599999999999

Total: 100 pts

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